Answer:
Freezing T° of solution = - 8.98°C
Explanation:
We apply Freezing point depression to solve this problem, the colligative property that has this formula:
Freezing T° of pure solvent - Freezing T° of solution = Kf . m
Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water
m = molality (moles of solute / 1kg of solvent)
We convert the mass of solvent from g to kg
1.5 g . 1kg/1000g = 0.0015 kg
We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g
0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles
Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m
- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent
-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C
Freezing T° of solution = - 8.98°C
Answer:
The temperature of the solution is -8.98 °C
Explanation:
Step 1: Data given
The normal freezing point of water is 0.00 ⁰C
Mass of ethylene glycol = 450.0 mg = 0.45 grams
Molar mass ethylene glycol = 62.07 g/mol
Mass of water = 1.5 grams = 0.0015 kg
Step 2: Calculate moles ethylene glycol
Moles ethylene glycol = mass / molar mass
Moles ethylene glycol = 0.45 grams / 62.07 g/mol
Moles ethylene glycol = 0.00725 moles
Step 3: Calculate molality
Molality = moles / mass
Molality = 0.00725 moles / 0.0015 kg
Molality = 4.83 molal
Step 4:
ΔT = i*Kf*m
⇒ i = the van't Hoff factor = 1
⇒ Kf = the freezing point depression constant = 1.86 °C/ m
⇒ m = molality = 4.83 molal
ΔT = 1 * 1.86 °C/m * 4.83 m
ΔT = 8.98
ΔT = T (pure solvent) − T (solution)
8.98 = 0.0 °C - (-8.98°C)
The temperature of the solution is -8.98 °C
